math
posted by APPRECIATIVE STUDENT .
Thank you for your help.
I don't know how to do this, so if you could explain step by step how to do this, that would be really good....
"A rough approximation for ln(5) is 1.609. Use this apporximation and differentials to approximate ln(521/100)
thanks

let y = ln(x)
x = 5
dy/dx = 1/x
dy = (1/x)dx
dx = 521/100  5 = .21 , y = ln(5) = 1.609
dy = (1/5)(.21) = .042
ln(521/100)
= ln (5.21) = appr. y + dy
= 1.609 + .042 = 1.651
check: ln(5.21) = 1.6505 (by calculator)
not bad
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